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            <h1 style="display: none">Educational Codeforces Round 109 Div. 2 2021.05.16</h1>
            
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              <h1 id="Educational-Codeforces-Round-109-Div-2-2021-05-16"><a href="#Educational-Codeforces-Round-109-Div-2-2021-05-16" class="headerlink" title="Educational Codeforces Round 109 Div. 2 2021.05.16"></a><a target="_blank" rel="noopener" href="https://codeforces.com/contest/1525">Educational Codeforces Round 109 Div. 2 2021.05.16</a></h1><h2 id="A-Potion-making"><a href="#A-Potion-making" class="headerlink" title="A. Potion-making"></a><a target="_blank" rel="noopener" href="https://codeforces.com/contest/1525/problem/A">A. Potion-making</a></h2><h3 id="题意"><a href="#题意" class="headerlink" title="题意"></a>题意</h3><p><img src="https://img-blog.csdnimg.cn/2021051618460769.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl81MDI4MTg2OQ==,size_16,color_FFFFFF,t_70" srcset="/img/loading.gif" lazyload alt="A"><br>两个数a，b占比分别为k%和(100-k)%，输入k求a+b最小值。</p>
<h3 id="题解"><a href="#题解" class="headerlink" title="题解"></a>题解</h3><p>不妨先取一个一定满足题意的，即a=k，b=100-k，然后再考虑缩小。分别除以gcd就好了。</p>
<h3 id="code"><a href="#code" class="headerlink" title="code"></a>code</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-keyword">using</span> <span class="hljs-keyword">namespace</span> std;<br><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">gcd</span><span class="hljs-params">(<span class="hljs-keyword">int</span> a,<span class="hljs-keyword">int</span> b)</span>   <span class="hljs-comment">//return a=gcd(a,b)</span></span><br><span class="hljs-function"></span>&#123;<br>    ll temp;<br>    <span class="hljs-keyword">while</span>(b)&#123;<br>        temp=b;<br>        b=a%b;<br>        a=temp;<br>    &#125;<br>    <span class="hljs-keyword">return</span> a;<br>&#125;<br><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span><br><span class="hljs-function"></span>&#123;<br>    <span class="hljs-keyword">int</span> ans=<span class="hljs-number">0</span>;<br>    <span class="hljs-keyword">int</span> t;cin&gt;&gt;t;<br>    <span class="hljs-keyword">while</span>(t--)&#123;<br>        ans=<span class="hljs-number">0</span>;<br>        <span class="hljs-keyword">int</span> a;cin&gt;&gt;a;<br>        <span class="hljs-keyword">int</span> b=<span class="hljs-number">100</span>-a;<br>        <span class="hljs-keyword">int</span> c=<span class="hljs-built_in">gcd</span>(a,b);<br>        cout&lt;&lt;a/c+b/c&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<br>    &#125;<br>    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;<br>&#125;<br></code></pre></td></tr></table></figure>

<h2 id="B-Permutation-Sort"><a href="#B-Permutation-Sort" class="headerlink" title="B. Permutation Sort"></a><a target="_blank" rel="noopener" href="https://codeforces.com/contest/1525/problem/B">B. Permutation Sort</a></h2><h3 id="题意-1"><a href="#题意-1" class="headerlink" title="题意"></a>题意</h3><p><img src="https://img-blog.csdnimg.cn/20210516185050500.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl81MDI4MTg2OQ==,size_16,color_FFFFFF,t_70" srcset="/img/loading.gif" lazyload alt="B"><br>给n和n个数，要求将其升序，每次可以选一个连续区间修改其中任意顺序，但是不能选一整个。</p>
<h3 id="题解-1"><a href="#题解-1" class="headerlink" title="题解"></a>题解</h3><p>特殊的地方就在首尾。分情况讨论即可。<br>如果1在第n个并且n在第一个，因为不能操作一整个区间，也就是说无法同时将这两个归位，所以需要通过两次操作把它们拿出来，然后再拿回去。答案就是3.<br>如果已经升序就是0.<br>如果首尾至少一个位置合法了，那么选择除了合法的之外的首尾那个区间进行操作即可，答案是1.<br>其它情况，也就是第一个和最后一个都放了其它数，两次操作分别把1和n放到合法位置，答案就是2.</p>
<h3 id="code-1"><a href="#code-1" class="headerlink" title="code"></a>code</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-keyword">const</span> <span class="hljs-keyword">int</span> N=<span class="hljs-number">55</span>;<br><span class="hljs-keyword">int</span> a[N];<br><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span><br><span class="hljs-function"></span>&#123;<br>    <span class="hljs-keyword">int</span> ans=<span class="hljs-number">0</span>;<br>    <span class="hljs-keyword">int</span> t;cin&gt;&gt;t;<br>    <span class="hljs-keyword">while</span>(t--)&#123;<br>        <span class="hljs-keyword">int</span> n;cin&gt;&gt;n;<br>        ans=<span class="hljs-number">0</span>;<br>        <span class="hljs-keyword">bool</span> yes=<span class="hljs-number">0</span>;<br>        <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> i=<span class="hljs-number">1</span>;i&lt;=n;i++)&#123;<br>            cin&gt;&gt;a[i];<br>            <span class="hljs-keyword">if</span>(i&gt;<span class="hljs-number">1</span>&amp;&amp;a[i]&lt;a[i<span class="hljs-number">-1</span>])yes=<span class="hljs-number">1</span>;<br>        &#125;<br>        <span class="hljs-keyword">if</span>(yes==<span class="hljs-number">0</span>)cout&lt;&lt;<span class="hljs-number">0</span>&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<br>        <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(a[<span class="hljs-number">1</span>]==n&amp;&amp;a[n]==<span class="hljs-number">1</span>)cout&lt;&lt;<span class="hljs-number">3</span>&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<br>        <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(a[<span class="hljs-number">1</span>]==<span class="hljs-number">1</span>||a[n]==n)cout&lt;&lt;<span class="hljs-number">1</span>&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<br>        <span class="hljs-keyword">else</span> cout&lt;&lt;<span class="hljs-number">2</span>&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<br>    &#125;<br>    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;<br>&#125;<br></code></pre></td></tr></table></figure>

<h2 id="D-Armchairs"><a href="#D-Armchairs" class="headerlink" title="D. Armchairs"></a><a target="_blank" rel="noopener" href="https://codeforces.com/contest/1525/problem/D">D. Armchairs</a></h2><h3 id="题意-2"><a href="#题意-2" class="headerlink" title="题意"></a>题意</h3><p><img src="https://img-blog.csdnimg.cn/20210516190603686.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl81MDI4MTg2OQ==,size_16,color_FFFFFF,t_70" srcset="/img/loading.gif" lazyload alt="D"><br>一开始有&lt;=(n/2)个1，把它们分别配对到一个0上，并且每个0只能配对一个1，花费为该1和该0的距离。要求将所有1移动到0的距离和最小值。</p>
<h3 id="题解-2"><a href="#题解-2" class="headerlink" title="题解"></a>题解</h3><p>dp一下。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><code class="hljs cpp">dp[i][j]:把前i个<span class="hljs-number">1</span>移动到前j个<span class="hljs-number">0</span>的最小花费<br>pos[p]:第p个<span class="hljs-number">1</span>或<span class="hljs-number">0</span>的下标（位置）<br>n1:<span class="hljs-number">1</span>数量<br>n0:<span class="hljs-number">0</span>数量<br>初始化dp:INF<br>从左往右枚举前i个<span class="hljs-number">1</span>移动的区间<br>同时这里枚举j是在[i,n0]内<br><span class="hljs-keyword">if</span> i==<span class="hljs-number">1</span> <br>   dp[i][j]=<span class="hljs-built_in">min</span>(dp[i][j<span class="hljs-number">-1</span>],<span class="hljs-built_in">abs</span>(pos[i]-pos[j])<br>   <span class="hljs-comment">//如果只有1一个1，那么就是在[1,j]区间里选一个位置使当前区间最小</span><br><span class="hljs-keyword">else</span> <br>   dp[i][j]=<span class="hljs-built_in">min</span>(dp[i][j<span class="hljs-number">-1</span>],dp[i<span class="hljs-number">-1</span>][j<span class="hljs-number">-1</span>]+<span class="hljs-built_in">abs</span>(pos[i]-pos[j])<br>   <span class="hljs-comment">//第一个转移：第i个1不放在新加入的第j个0处，那么就要把i个1放在前j-1个0，特殊地，如果不够放即j-1&lt;i，因为上述初始化为INF，所以取min时默认后一个转移选项</span><br>   <span class="hljs-comment">//第二个转移：第i个1放在新加入的第j个0处，那么当前费用就是前i-1个1放在前j-1个0处加上第i个1放在第j个0处的费用。</span><br></code></pre></td></tr></table></figure>

<p>解释一下为什么j左界是i，因为我们这里i是从左往右数的，数到i和i+1个，对于第i个0，把i+1移动到那和把i移动到那明显后者更优，也就是说宁可移动i也不需要移动i+1，前面的也同理。所以不需要考虑i前面的0。同时也是为了能使每个i有地方放（这点在初始化与取min时已经排除了影响，实际上可有可无）。</p>
<h3 id="code-2"><a href="#code-2" class="headerlink" title="code"></a>code</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-keyword">using</span> <span class="hljs-keyword">namespace</span> std;<br><span class="hljs-keyword">int</span> n1,n0;<br><span class="hljs-keyword">const</span> <span class="hljs-keyword">int</span> N=<span class="hljs-number">5001</span>;<br><span class="hljs-keyword">int</span> c1[N],c0[N];<br><span class="hljs-keyword">int</span> dp[N][N];<br><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span><br><span class="hljs-function"></span>&#123;<br>    <span class="hljs-keyword">int</span> n;cin&gt;&gt;n;<br>    <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> i=<span class="hljs-number">1</span>;i&lt;=n;i++)&#123;<br>        <span class="hljs-keyword">int</span> d;cin&gt;&gt;d;<br>        <span class="hljs-keyword">if</span>(d==<span class="hljs-number">1</span>)&#123;<br>            c1[++n1]=i;<br>        &#125;<br>        <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(d==<span class="hljs-number">0</span>)&#123;<br>            c0[++n0]=i;<br>        &#125;<br>    &#125;<br>    <span class="hljs-keyword">if</span>(n1==<span class="hljs-number">0</span>)&#123;cout&lt;&lt;<span class="hljs-number">0</span>&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;&#125;<br>    <span class="hljs-built_in">memset</span>(dp,<span class="hljs-number">0x3f</span>,<span class="hljs-keyword">sizeof</span> dp);<br>    <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> i=<span class="hljs-number">1</span>;i&lt;=n1;i++)&#123;<br>        <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> j=i;j&lt;=n0;j++)&#123;<br>            <span class="hljs-keyword">if</span>(i==<span class="hljs-number">1</span>)dp[i][j]=<span class="hljs-built_in">min</span>(dp[i][j<span class="hljs-number">-1</span>],<span class="hljs-built_in">abs</span>(c1[i]-c0[j]));<br>            <span class="hljs-keyword">else</span> dp[i][j]=<span class="hljs-built_in">min</span>(dp[i][j<span class="hljs-number">-1</span>],dp[i<span class="hljs-number">-1</span>][j<span class="hljs-number">-1</span>]+<span class="hljs-built_in">abs</span>(c1[i]-c0[j]));<br>        &#125;<br>    &#125;<br>    cout&lt;&lt;dp[n1][n0]&lt;&lt;<span class="hljs-string">&#x27;\n&#x27;</span>;<br>    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;<br>&#125;<br><br></code></pre></td></tr></table></figure>


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